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\begin{document}

\title{高等代数二}
\subtitle{03-重因式-多项式函数-多项式的根}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月7日} }

\maketitle

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\begin{frame}{3.1. 作业：星期天晚上十点半之前在网络教学平台提交 }

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\begin{enumerate}
\item   整理课堂笔记里的重点难点。补充没写完的计算或证明。
\item   习题(2.5)任选2题。抄写题目。
\item   习题(2.6)任选2题。抄写题目。
\end{enumerate}

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\begin{enumerate}

\item   单因式、重因式
\item   多项式的导数多项式
\item   判断一个多项式是否存在重因式

\item   多项式函数
\item   余式定理
\item   综合除法和长除法
\item   多项式的根
\item   拉格朗日插值公式

\end{enumerate}

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\begin{enumerate}

\item   使用导数多项式来计算多项式的重因式。
\item   余式定理的含义与证明。
\item   多项式的一次因式与多项式函数的根之间的关系。
\item   拉格朗日插值公式。
%\item   

\end{enumerate}

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\begin{itemize}

\item  {\color{red}问：什么是单因式？}

\item 答：设 $p(x)$ 是一个不可约多项式。若 $p(x)\mid f(x)$ 但 $p(x)^2\nmid f(x)$, 则称 $p(x)$ 是 $f(x)$ 的一个单因式。


\item  {\color{red}问：什么是重因式？}

\item 答：设 $p(x)$ 是一个不可约多项式。若 $p(x)^k\mid f(x)$ 但 $p(x)^{k+1}\nmid f(x)$, 其中 $k\ge 2$, 则称 $p(x)$ 是 $f(x)$ 的一个 $k$ 重因式。


\end{itemize}

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\begin{itemize}

%\item 注：单因式即1重因式。

\item  {\color{red}问题：求实数域上的多项式 $f(x)=(x+1)(x^2+2)^2$ 的单因式和重因式。}

\item  解答：
\begin{enumerate}
\item  多项式 $x+1$ 是 $f(x)$ 的单因式。
\item  多项式 $x^2+2$ 是 $f(x)$ 的2重因式。
\end{enumerate}


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：求实数域上的多项式 $f(x)=(x+1)(x^2-2)^2$ 的单因式和重因式。}

\item  解答：
\begin{enumerate}
\item  多项式 $x+1$ 是 $f(x)$ 的单因式。
\item  多项式 $x-\sqrt{2}$ 是 $f(x)$ 的2重因式。
\item  多项式 $x+\sqrt{2}$ 是 $f(x)$ 的2重因式。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：求有理数域上的多项式 $f(x)=(x+1)(x^2-2)^2$ 的单因式和重因式。}

\item  解答：
\begin{enumerate}
\item  多项式 $x+1$ 是 $f(x)$ 的单因式。
\item  多项式 $x^2-2$ 是 $f(x)$ 的2重因式。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item A: 虽然我们还没有一般的方法来分解因式，但是我们竟然有一般的方法来判断一个多项式有没有重因式。

\item B: 真是一件神奇的事情。你是怎么做到的呢？

\item A: 使用导数多项式与辗转相除法。

\end{itemize}

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\begin{itemize}

\item  {\color{red}什么是多项式的导数多项式？}

\item  定义：一个多项式 $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ 的导数多项式是指
$$f\,'(x):= a_1+2a_2x+\cdots+na_nx^{n-1}.$$

\item  注：这里的导数是只针对多项式来定义的，是形式化定义的，跟极限无关。就是给上式右边的表达式起个名字，把它称为 $f(x)$ 的导数多项式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}多项式的导数运算具有下述性质：

设 $f(x),g(x)\in F[x]$, 则有线性性质、莱布尼兹公式、幂求导公式成立，
\begin{eqnarray*}
(kf(x)+\ell g(x))\,' &=& kf\,'(x) + \ell g\,'(x), \\ 
(f(x)g(x))\,' &=& f\,'(x)g(x) + f(x)g\,'(x), \\
(f(x)^k)\,' &=& k(f(x))^{k-1}f\,'(x).
\end{eqnarray*}
}

\item  证明：先证明线性性质。再证明单项式的情形。再证明一般的多项式。或者将多项式的一般形式具体写出来，两边直接计算。

\end{itemize}

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\begin{itemize}

\item {\color{red}定理2.5.1. 设 $p(x)$ 是一个不可约多项式。}
\begin{enumerate}
\item  {\color{red}若 $p(x)$ 是 $f(x)$ 的一个 $2$ 重因式，则 $p(x)$ 是 $f\,'(x)$ 的单因式。}
\item  {\color{red}若 $p(x)$ 是 $f(x)$ 的一个 $k$ 重因式$(k\ge 3)$, 则 $p(x)$ 是 $f\,'(x)$ 的 $k-1$ 重因式。}
\end{enumerate}


\item 证明：先写出下述计算，
\begin{eqnarray*}
f(x) &=& p(x)^kg(x), \\
f\,'(x) &=& kp(x)^{k-1}p\,'(x)g(x) + p(x)^kg\,'(x) \\ 
&=& p(x)^{k-1}[kp\,'(x)g(x) + p(x)g\,'(x)]. 
\end{eqnarray*}
然后证明 $p(x)$ 不能整除中括号里这项。
\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.5.2. 下述两个条件是等价的。}
\begin{enumerate}
\item[(i)]  {\color{red}多项式 $f(x)$ 没有重因式。}
\item[(ii)] {\color{red}多项式 $f(x)$ 与其导数多项式 $f\,'(x)$ 互素。}
\end{enumerate}

\item  例子：下述 $f(x)$ 有重因式，
\begin{eqnarray*}
f(x) &=& (x-1)^3, \\
f\,'(x) &=& 3(x-1)^2.
\end{eqnarray*}
%则 $f(x)$ 与 $f\,'(x)$ 不是互素的。 

\item  例子：下述 $f(x)$ 没有重因式，
\begin{eqnarray*}
f(x) &=& (x-1)(x-2)=x^2-3x+2, \\
f\,'(x) &=& 2x-3.
\end{eqnarray*}
%则  $f(x)$ 与 $f\,'(x)$ 是互素的。

\end{itemize}

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\begin{itemize}

\item {\color{red}证明定理2.5.2. }

\begin{enumerate}

\item  先将 $f(x)$ 写出典型分解式
$$f(x)=ap_1(x)^{k_1}p_2(x)^{k_2}\cdots p_t(x)^{k_t}.$$

\item  计算 $f\,'(x)$. 经过一番测试(设 $t=1$ 和 $t=2$)，可得
$$f\,'(x)=p_1(x)^{k_1-1}p_2(x)^{k_2-1}\cdots p_t(x)^{k_t-1}g(x),$$
其中 $g(x)$ 没有 $p_1(x),p_2(x),\cdots,p_t(x)$ 这些因式。

\item  于是当 $k_1=k_2=\cdots=k_t=1$ 时，$f(x)$ 与 $f\,'(x)$ 互素。

\item  如果 $f(x)$ 有重因式，那么这个不可约因式仍会在 $f\,'(x)$ 中出现。

\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：如何判断一个多项式是否有重因式？}

\item  答：用辗转相除法计算 $f(x)$ 与 $f\,'(x)$ 的最大公因式。若 $f(x)$ 与 $f\,'(x)$ 互素，则 $f(x)$ 没有重因式。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $p_1(x)$ 与 $p_2(x)$ 是互素的两个不可约多项式，设有
\begin{eqnarray*}
f(x) = p_1(x)^3p_2(x)^4. 
\end{eqnarray*}
求 $f(x)$ 与 $f\,'(x)$ 的最大公因式，并由此求 $f(x)$ 的重因式。}

\item  解答：多项式 $f(x)$ 的导数多项式为 
\begin{align*}
f\,'(x) &= 3p_1(x)^2p_1\,'(x)p_2(x)^4 + 4p_1(x)^3p_2(x)^3p_2\,'(x), \\ 
 &= p_1(x)^2p_2(x)^3[3p_1\,'(x)p_2(x) + 4p_1(x)p_2\,'(x) ]
\end{align*}
因此 $f(x)$ 与 $f\,'(x)$ 的最大公因式为 $p_1(x)^2p_2(x)^3$.
从而推知 $f(x)$ 有3重因式 $p_1(x)$ 与4重因式 $p_2(x)$. 

%\item  注：实际情况中，多项式可能不是预先就因式分解好的。求出 $(f,f\,')$ 之后，还需要因式分解，才能找出重因式。所以这个方法只能判断有没有重因式，不一定能找出重因式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：如何把一个多项式看作一个函数？}

\item  答：设 $f(x)=a_0+a_1x+\cdots+a_nx^n$ 是系数在数环 $R$ 中的一个多项式。则当 $x$ 在 $R$ 中取值的时候，$f(x)$ 也将在 $R$ 中取值。于是 
$$f : R\to R$$ 
成为一个函数。这称为多项式函数。

\item  注：因为数环中的数在加法、减法和乘法下的运算结果是封闭的，所以对任意 $c\in R$, 有 $f(c)\in R$. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.6.1. 设 $R$ 是一个数环，设 $f(x)\in R[x]$, 设 $c\in R$. 则 $f(x)$ 除以 $(x-c)$ 得到的余式正好是多项式函数 $f(x)$ 在 $x=c$ 时的函数值 $f(c)$.}

\item  证明：根据带余除法，有 $$f(x)=(x-c)q(x)+r.$$
代入 $x=c$, 可得 $f(c)=r$. 

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是综合除法？什么是长除法？}

\begin{figure}
\centering
\includegraphics[height=0.6\textheight, width=0.8\textwidth]{pic/synthetic-division.jpg}
% \caption{ }
\end{figure}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：用 $x+3$ 除 $f(x)=x^4+x^2+4x-9$, 求得到的商和余式。 } 

\item  解答：商式 $q(x)=x^3-3x^2+10x-26$, 余式 $r=69$. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}定义：什么是多项式的根？}

\item  答：设 $R$ 是一个数环，设 $f(x)\in R[x]$, 设 $c\in R$. 若 $f(c)=0$, 则称 $c$ 是 $f(x)$ 在 $R$ 中的一个根。

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\begin{itemize}

\item  {\color{red}定理2.6.2. 数 $c$ 是 $f(x)$ 的根的充分必要条件是 $x-c$ 整除 $f(x)$. } 

\item  证明：根据带余除法，存在多项式 $q(x)$ 与 $r$ 使得 $$f(x)=(x-c)q(x)+r. $$ 于是下述三个条件是等价的：
\begin{enumerate}
\item  $f(c)=0$.  
\item  $r=0$.  
\item  $(x-c)\mid f(x)$. 
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.6.3. 设 $n\ge 0$. 设 $R$ 是一个数环。设 $f(x)$ 是 $R[x]$ 中的一个 $n$ 次多项式。则 $f(x)$ 在 $R$ 中至多有 $n$ 个根。}

\item  证明：对 $n$ 进行归纳法。
\begin{enumerate}
\item  当 $n=0$ 时，$f(x)$ 是零次多项式，所以 $f(x)=c\neq 0$, 所以 $f(x)$ 没有根。
\item  归纳假设：设已证 $n$ 次多项式至多有 $n$ 个根。
\item  现设 $f(x)$ 是一个 $n+1$ 次多项式。设 $c\in R$ 是 $f(x)$ 的一个根。则 $f(x)=(x-c)g(x)$, 其中 $g(x)$ 为一个 $n$ 次多项式。
\item  由归纳假设，$g(x)$ 至多有 $n$ 个根。
\item  所以 $f(x)$ 至多有 $n+1$ 个根。
\end{enumerate}


%\item  问题：有没有可能，存在 $d\in R$, 使得 $d-c\neq 0$ 且 $g(d)\neq 0$, 但是 $$f(d)=(d-c)g(d)=0 ?$$


\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.6.4. 设 $f(x),g(x)\in R[x]$ 是两个次数都小于等于 $n$ 的多项式。设存在 $n+1$ 个不同的数 $c_1,c_2,\cdots,c_{n+1}\in R$
使得 $$f(c_i)=g(c_i),\,\, i=1,2,\cdots, n+1, $$ 则 $f(x)=g(x)$, 即这是两个相同的多项式。}

\item  证明：根据条件，次数至多为 $n$ 的多项式 $u(x)=f(x)-g(x)$ 在 $R$ 中有了 $n+1$ 个不同的根。由上一个定理，这个多项式只能是零多项式。

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.6.5. 设 $f(x),g(x)\in R[x]$ 是两个多项式。则它们是同一个多项式当且仅当它们作为 $R\to R$ 的函数是同一个函数。}

\item  证明思路：
\begin{enumerate}
\item  作为多项式相等，是指系数对应相等。
\item  作为函数相等，是指对任意 $c\in R$, 有 $f(c)=g(c)$. 
\item  因为数环 $R$ 中有无穷多个数，根据上一个定理即得证明。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是拉格朗日插值公式？}

\item  答：给定数域 $F$ 中的一些数 $a_i,b_i, 1\le i\le n$, 其中 $a_i$ 互不相同。则存在一个次数不超过 $n$ 的多项式 $f(x)$, 使得 $f(a_i)=b_i, 1\le i\le n$. 拉格朗日插值公式就是这个问题的解答。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：给定 $a_1\neq a_2$ 以及任意的 $b_1,b_2$, 求一次多项式 $f(x)$ 使得 $$f(a_i)=b_i,\quad i=1,2.$$
}

\item  解答：下述公式称为拉格朗日插值公式，
$$f(x) = \frac{b_1(x-a_2)}{a_1-a_2} + \frac{b_2(x-a_1)}{a_2-a_1}.$$

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\begin{itemize}

\item  {\color{red}问题：给定互不相同的三个数 $a_1, a_2, a_3$ 以及任意的 $b_1,b_2,b_3$, 求次数不超过2的多项式 $f(x)$ 使得 $$f(a_i)=b_i,\quad i=1,2,3.$$
}

\item  解答：下述公式称为拉格朗日插值公式，
$$f(x) = \frac{b_1(x-a_2)(x-a_3)}{(a_1-a_2)(a_1-a_3)} + \frac{b_2(x-a_1)(x-a_3)}{(a_2-a_1)(a_2-a_3)} + \frac{b_3(x-a_1)(x-a_2)}{(a_3-a_1)(a_3-a_2)}.$$

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\begin{itemize}

\item  {\color{red}问题：求次数不超过2的多项式 $f(x)$ 使得 $$f(1)=1,\,\, f(-1)=3,\,\, f(2)=3. $$ }

\item  解答：根据拉格朗日插值公式，计算可得 $f(x)=x^2-x+1$. 

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\begin{itemize}

\item   {\color{red}问题：证明下述关于多项式的导数的等式，
\begin{enumerate}
\item  {\color{red} $(f(x)+g(x))' = f\,'(x) + g\,'(x)$. } 
\item  {\color{red} $(f(x)g(x))' = f\,'(x)g(x) + f(x)g\,'(x)$. } 
\end{enumerate}
}

\item  解答：将多项式具体写出来，按导数的定义验证。


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\begin{itemize}

\item  {\color{red}问题：设有正整数 $k\ge 3$, 设 $p(x)$ 是 $f\,'(x)$ 的 $k-1$ 重因式，并设 $p(x)$ 整除 $f(x)$. 证明 $p(x)$ 是 $f(x)$ 的 $k$ 重因式。} 

\item  解答：使用重因式的定义。设 $f(x)=p(x)^mg(x)$, 其中 $p(x)\nmid g(x)$, 证明 $m=k$. 


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\begin{itemize}

\item  {\color{red}问题：证明下述有理系数多项式没有重因式：$$f(x)=1+x+\frac{x^2}{2!} + \cdots + \frac{x^n}{n!}. $$
}

\item  解答：计算 $f\,'(x)$ 与 $f(x)$ 的最大公因式。

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\begin{itemize}

\item   {\color{red}问题：设 $a,b$ 是两个有理数。设 $f(x)$ 有重因式。求 $a,b$ 所满足的条件。}
\begin{enumerate}
\item  {\color{red} $f(x)=x^3+3ax+b$. } 
\item  {\color{red} $f(x)=x^4+4ax+b$. } 
\end{enumerate}

\item  解答：应用多项式有重因式的充分必要条件。
\begin{enumerate}
\item  $4a^3+b^2=0$.  
\item  $27a^4-b^3=0$. 
\end{enumerate}

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\begin{itemize}

\item   {\color{red}问题：设 $F$ 是一个数域，设 $f(x)\in F[x]$ 是一个 $n$ 次多项式。证明下述条件等价。}
\begin{enumerate}
\item[(i)]  {\color{red} $f\,'(x)$ 整除 $f(x)$.  } 
\item[(ii)]  {\color{red} 存在 $a,b\in F, a\neq 0$ 使得 $f(x) = a(x-b)^n$. } 
\end{enumerate}

\item  解答：
\begin{enumerate}
\item  (ii) $\Rightarrow$ (i). 直接计算导函数。
\item  (i) $\Rightarrow$ (ii). 考虑 $f(x)$ 的典型分解式。
\end{enumerate}

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\begin{itemize}

\item  {\color{red}问题：设 $f(x)=2x^5-3x^4-5x^3+1$, 求 $f(3)$ 与 $f(-2)$. }

\item  解答：有几种方法。
\begin{enumerate}
\item 直接代入计算。
\item 用长除法或综合除法，计算 $f(x)$ 除以 $x-3$ 的余式，余式为 $f(3)$. 
\end{enumerate}
\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设下述等式成立，求 $a,b,c,d$ 的值，
\[ 2x^3-x^2+3x-5 = a(x-2)^3 +b(x-2)^2 +c(x-2) +d. \]
}

\item  解答：有几种方法。
\begin{enumerate}
\item 代入 $x=2$ 可得 $d$. 两边求导，再代入 $x=2$ 可得 $c$. 
\item 将 $f(x)=2x^3-x^2+3x-5$ 除以 $g(x)=x-2$, 商为 $q(x)$, 余式为 $d$. 将 $q(x)$ 再除以 $g(x)=x-2$, 余式为 $c$. 
%\item 
\end{enumerate}

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\begin{itemize}

\item  {\color{red}问题：求一个次数小于 4 的多项式 $f(x)$ 使得 
\begin{align*}
f(2)=3, f(3)=-1, f(4)=0, f(5)=2. 
\end{align*}
}
\item  解答：使用拉格朗日插值公式可得
$$f(x)=-\frac{2}{3}x^3 + \frac{17}{2}x^2 - \frac{203}{6}x + 42. $$

\item  使用R程序验证：
\begin{lstlisting}[language=R]
f = function(x) -2*x^3/3 + 17*x^2/2 -203*x/6 +42
f(2); f(3); f(4); f(5)
f(c(2,3,4,5))
\end{lstlisting}

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\begin{itemize}

\item  {\color{red}问题：设 $f(x)$ 和 $g(x)$ 是两个多项式，设多项式 $x^2+x+1$ 整除多项式 $f(x^3)+xg(x^3)$. 证明 $f(1)=g(1)=0$. }

\item  解答：考虑 $x^2+x+1$ 的两个复根。

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\begin{itemize}

\item  {\color{red}问题：设 $f(x)$ 是一个复系数多项式，设 $f(x)\neq 0$. 设 $f(x)\mid f(x^3)$. 证明 $f(x)$ 的根只能是零或者单位根，即 $x^m=1$ 的复根。}

\item  解答：若 $f(c)=0$ 则 $f(c^3)=0, f(c^6)=0, f(c^9)=0$. 

\end{itemize}

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